Chapter 10.2, Problem 71E

To determine

To find: The mistake in the method used by Jannie to construct the confidence interval.

Answer to Problem 71E

Jannie has constructed confidence interval for difference in two sample proportions instead of one-sample proportion.

Explanation of Solution

Given:

Summary statistic is:

  

Formula used:

The formula to construct the confidence interval for one-sample proportion is:

  $\widehat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

Calculation:

Here, it is required to compute the confidence interval for one-sample proportion instead of difference in proportions. The mistake done by Jannie is that she has computed confidence interval for difference in proportions instead of one-sample proportion.

The calculation for the required confidence interval could be done as:

The sample proportion is calculated as:

  $\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{427}{427+2733}\\ =0.135\end{array}$

The z-score at 99% confidence level is 2.576.

The confidence interval is:

  $\begin{array}{c}CI=\widehat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =0.135\pm 2.576\times \sqrt{\frac{0.135\left(1-0.135\right)}{3160}}\\ =\left(0.119,0.151\right)\end{array}$

Thus, the required confidence interval is $\left(0.119,0.151\right)$ .

Interpretation:

There is 99% probability that the proportion of student that are taking the SAT twice are being coaches lies between 0.119 and 0.151.